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It is mentioned in the introduction of the paper VB-GROUPOIDS AND REPRESENTATION THEORY OF LIE GROUPOIDS by LFONSO GRACIA-SAZ AND RAJAN AMIT MEHTA https://arxiv.org/pdf/1007.3658.pdf that there is a one-one correspondence between the set of Lie group representations on finite dimensional vector spaces and the set of group objects in the category of smooth vector bundles.

In one direction they said that given a Lie group $G$ and a representation $\rho: G \rightarrow \rm{Aut}(V)$ on a finite dimensional vector space $V$, the semi direct product $G \rtimes_{\rho} V \rightarrow G$ is a group object in the category of smooth vector bundles.

In the converse direction, they said that any group object $\pi: E \rightarrow G$ in the category of smooth vector bundles induce a Lie group structure on $G$ and a representation $\mu: G \rightarrow \rm{Aut}(\pi^{-1}(e))$ such that $E$ can be identified with $G \rtimes_{\mu} \pi^{-1}(e)$.

My confusion is the following: According to the answer to the question Product in the category of smooth vector bundles given by user @Thorgott, the multiplication map associated to the group object $\pi:E \rightarrow G$ is a morphism of vector bundles ($m_E:E \times E \rightarrow E$, $m_G : G \times G \rightarrow G$) such that it behaves nicely with identity and inverse maps. I am assuming that they meant to say that $m_G:G \times G \rightarrow G$ is the map which gives a canonical Lie group structure on the manifold $G$. But, then what is the canonical representation $\mu:G \rightarrow \rm{Aut}(\pi^{-1}(e))$ they are considering here? They mentioned about right translations of zero vectors but I am not able to guess the map $\mu$ properly!

Edit: Are they referring about the following map?

For each $g \in G$, $\mu_g:\pi^{-1}(e)\rightarrow \pi^{-1}(e)$ is defined as $p \mapsto p.0_g \mapsto \phi_{g}(p.0_g)\mapsto \phi_e^{-1}(\phi_{g}(p.0_g))$ where $\phi_g: \pi^{-1}(U_g) \rightarrow U_g \times \mathbb{R}^{n}$ and $\phi_e: \pi^{-1}(U_e) \rightarrow U_e \times \mathbb{R}^{n}$ are local trivialisations around $g$ and $e$ respectively and $n$ is the rank of the vector bundle $\pi:E \rightarrow G$.

Am I miss understanding something here?

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    $\begingroup$ Did you try to identify the fibre $E_g$ with $E_e$ via the right translation map of $E$ $R^E_{0_{g^{-1}}}:E_g \to E_e, v\mapsto v\cdot 0_{g^{-1}}$? Maybe this will give you the trivialization they refer to. $\endgroup$
    – T.P.
    Dec 3, 2022 at 17:16
  • $\begingroup$ @T.P. Yes, I considered the map in the version, $E_e \rightarrow E_g$ taking $p \mapsto p.0_g$. May be I am missing some thing very trivial but I am not able to produce the next step i.e for each $g \in G$, an isomorphism from $E_e$ to $E_e$. $\endgroup$ Dec 3, 2022 at 17:24
  • $\begingroup$ @T.P. Please check the edit I made. Is it making sense? $\endgroup$ Dec 3, 2022 at 18:02
  • $\begingroup$ What if $U_g$ and $U_e$ are disconnected? I think there is something wrong with your construction. Maybe the automorphism of $E_e$ is given by some sort of conjugation: $E_e \to E_e, v\mapsto 0_{g^{-1}}v0_{g}$? Try if something like this would work. $\endgroup$
    – T.P.
    Dec 3, 2022 at 21:19

2 Answers 2

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I post this as an answer because the discussion in the comments section is getting too long :D


The canonical identification of the fiber $E_g$ with $E_e$ is given, as you said, by right translations of zero vectors: for each $g\in G$, $R_{0_{g^{-1}}}:E_g\to E_e$ is defined by $R_{0_{g^{-1}}}(v)=v\cdot 0_{g^{-1}}$.

Your representation $\mu$ should send an element $g\in G$ to the automorphism given by left are right conjugation of zero vectors; that is, $\mu(g)$ is the following composition: $$ L_{0_{g}}\circ R_{0_{g^{-1}}}:E_e\to E_g\to E_e;\ v\mapsto 0_{g}\cdot v \cdot 0_{g^{-1}}$$ Note that this is a reasonable idea since conjugation appears in the usual (inner) semidirect product of groups (you can also check the Wikipedia page).

Now due to the interchange law of VB-Group(oid)s (check Definition 3.1, part (4) in the paper), we have that $$ 0_{g}\cdot 0_{h} = (0_{g} + 0_{g})\cdot (0_{h} + 0_{h}) = 0_{g}\cdot 0_{h} + 0_{g}\cdot 0_{h} $$ and therefore $0_{g}\cdot 0_{h} = 0_{gh}$; this also is to be expected since the group structure is compatible with the vector bundle operations. This proves that the map $\mu$ defined above is indeed a group homomorphism and hence a representation of $G$ on $E_e$.

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  • $\begingroup$ Thank you for the answer. It seems local trivialisations may not play any role for the representation. $\endgroup$ Dec 5, 2022 at 3:51
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    $\begingroup$ I’d say that on that level of differential geometry, and in particular in graded geometry and higher structures (i.e. VB objects, graded manifolds, $L_\infty$ objects), the majority of constructions do not use charts and trivializations. Most of the stuff appearing here are expressed in a canonical way. $\endgroup$
    – T.P.
    Dec 5, 2022 at 8:25
  • $\begingroup$ Thanks for the clarification! $\endgroup$ Dec 6, 2022 at 12:58
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Though, the answer by @T.P.looks completely fine to me, but I think I observed something more general in the one-one correspondence between the set of Lie group representations on finite dimensional vector spaces and the set of group objects in the category of smooth vector bundles, which I thought of sharing in the forum. Please correct me if my observation does not make much sense!

So, given a group object $\pi : E \rightarrow G$ in the category of smooth vector bundles, we need to produce a representation $\mu:G \rightarrow \rm{Aut}(\pi^{-1}(e))$.

From the fact that $\pi : E \rightarrow G$ is a group object, we can represent $G= [G \rightrightarrows *]$ and $E=[E \rightrightarrows *]$ as groupoids. Now if represent the vector bundle $\pi :E \rightarrow G$ as a VB-groupioid $(\pi,\rm{id}):[E \rightrightarrows *] \rightrightarrows [G \rightrightarrows *]$ then, it is not difficult to show that $[E \rightrightarrows *]$ is a fibered category over the groupoid $[G \rightrightarrows*]$. Let us choose the zero section of the vector bundle $\pi: E \rightarrow G$ as a cleavage for the fibered category. Now, we construct the associated pseudofunctor(which in our case is actually a contravariant functor!) $F:[G \rightrightarrows*] \rightarrow \rm{Cat}$, then $F(*)=[\pi^{-1}(e)\rightrightarrows \rm{id}^{-1}(*)])$, and for each $g \in G$ we get a map $F(g): \pi^{-1}(e) \rightarrow \pi^{-1}(e)$. [I think it is possible to show that $F(g)$ is a linear map with respect to the vector space structure of $\pi^{-1}(e)$].Finally, by the functoriality of $F$ we get a group homomorphism $F:G \rightarrow \rm{Aut}(\pi^{-1}(e))$ which is actually the required representation of $G$ associated to the vector bundle $\pi :E \rightarrow G$. So, $\mu$ is $F$.

[Though I didn't talk about smoothness but I think it will not make much trouble if we incorporate it in the set up]

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